Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

For inequality questions we're not going to simply substitute all the values, but we'll be comparing the range of the original question and the range of conditions and investigate if the range of the question includes that of conditions. If it does, then the condition is sufficient.

anartey wrote:

Is \(k^2 + k - 2 > 0\)?

(1) \(k < 1\)

(2) \(k < -2\)

The first step is simplifying the original question.

\(k^2 + k - 2 > 0\)

\((k+2)(k-1)>0\)

\(k < -2\) or \(k > 1\)

Condition (1)

The range of the question, "\(k < -2\) or \(k > 1\)" does not include the range of condition (1), "\(k < 1\)".

Thus this is NOT sufficient.

Condition (2)

The range of the question, "\(k < -2\) or \(k > 1\)" includes the range of condition (2), "\(k < -2\)".

Normally for cases where we need 1 more equation, such as original conditions with "1 variable", or "2 variables and 1 equation", or "3 variables and 2 equations", we have 1 equation each in both 1) and 2). Therefore D has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) separately. Here, there is 59 % chance that D is the answer, while A or B has 38% chance. There is 3% chance that C or E is the answer for the case. Since D is most likely to be the answer according to DS definition, we solve the question assuming D would be our answer hence using 1) and 2) separately. Obviously there may be cases where the answer is A, B, C or E.

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